Chemistry - Chemical Kinetics Question with Solution | TestHub

Given, initial temperature, $T_1=20+273=293\mathrm{K}$ Final temperature $\begin{array}{rl}{T}_2& =35+273\\ & =308\mathrm{K}\\ R& =8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\end{array}$ Since, rate becomes double on raising temperature, $\therefore r_2=2r_1\text{ or }\frac{r_2}{r_1}=2$ As rate constant, $k\propto r$ $\therefore \frac{k_2}{k_1}=2$ From Arrnhenius equation, we know that $\begin{array}{rl}& \log \frac{k_2}{k_1}=-\frac{E_2}{2.303R}\left[\frac{T_1-T_2}{T_1{T}_2}\right]\\ & \log 2=-\frac{E_2}{2.303\times 8.314}\left[\frac{293-308}{293\times 308}\right]\\ & 0.3010=-\frac{E_2}{2.303\times 8.314}\left[\frac{-15}{293\times 308}\right]\end{array}$ $\begin{array}{rl}\therefore E_2& =\frac{0.3010\times 2.303\times 8.314\times 293\times 308}{15}\\ & =34673.48\mathrm{J}{\mathrm{mol}}^{-1}=34.7\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$